Question: Simplify; express your answer in exponential form. Assume $r\neq 0, y\neq 0$. $\dfrac{{(r^{-2}y^{-1})^{5}}}{{(r^{-3}y)^{-3}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(r^{-2}y^{-1})^{5} = (r^{-2})^{5}(y^{-1})^{5}}$ On the left, we have ${r^{-2}}$ to the exponent ${5}$ . Now ${-2 \times 5 = -10}$ , so ${(r^{-2})^{5} = r^{-10}}$ Apply the ideas above to simplify the equation. $\dfrac{{(r^{-2}y^{-1})^{5}}}{{(r^{-3}y)^{-3}}} = \dfrac{{r^{-10}y^{-5}}}{{r^{9}y^{-3}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-10}y^{-5}}}{{r^{9}y^{-3}}} = \dfrac{{r^{-10}}}{{r^{9}}} \cdot \dfrac{{y^{-5}}}{{y^{-3}}} = r^{{-10} - {9}} \cdot y^{{-5} - {(-3)}} = r^{-19}y^{-2}$